Question: $ f(x)=\sum_{n=0}^{\infty }\dfrac{n+2}{2^{n}}x^{n+1}$ $\int_{0}^{1}f(x)dx=$
First, perform the requested integration. $\begin{aligned} &\phantom{=}\int\limits_{0}^{1}{f\left( x \right)dx} \\\\ &=\int_0^1\sum_{n=0}^{\infty }\dfrac{n+2}{2^{n}}x^{n+1}dx \\\\ &=\sum_{n=0}^{\infty }\int_0^1\dfrac{n+2}{2^{n}}x^{n+1}dx \\\\ &=\left. \sum_{n=0}^{\infty }\dfrac{n+2}{2^{n}}\dfrac{x^{n+2}}{n+2} \right|_{0}^{1} \\\\ &=\sum_{n=0}^{\infty }{\dfrac{1}{{{2}^n}}} \end{aligned}$ Now recognize that the result is a geometric series with first term $1$ and common ratio $\dfrac{1}{2}$. Hence, the series converges to $\dfrac{1}{1-\dfrac{1}{2}}=2$.